Is the code after 'break' executed?

Question

for (...)
    for (...)
    {
        break;
        break;   // 1
    }

Will the code at (1) execute? There could also be there continue or anything else. I know I could just check in my debugger but I wanna know what C++ standard says about it as my compiler might just be a special case.

"compilers don't do such "special" thing " - Of course, they do. Various compiler producers add their own extensions/improvements to/beyond the language standard. Sometimes because they think it's better than what standard offers, sometimes because it's just easier for them. I assume. ** Of course, the differences aren't huge but they are there. — NPS, Aug 07 '13 at 10:36
@JBL I'm pretty sure I've seen this construct in some other language (Java/C#/something) where specifically double `break` is allowed and breaks 2 nested `for`s. — NPS, Aug 07 '13 at 10:49
@NPS Strange, but in this case, that seems like something the compiler interprets in its own way, because in this case, the behavior of the statement wouldn't be consistent... In The New Idiot's answer, the standard for C++ makes it pretty clear anyway that control passes immediately to the next statement out of the scope `break` just exited. — JBL, Aug 07 '13 at 10:57

score 1 · Accepted Answer · answered Aug 07 '13 at 10:32

1

As per the specifications

6.6.1 The break statement [stmt.break]

The break statement shall occur only in an iteration-statement or a switch statement and causes termination of the smallest enclosing iteration-statement or switch statement; control passes to the statement following the terminated statement, if any.

Hence 1 should not even reach . Some Java compiler might even flag that code unreachable.

answered Aug 07 '13 at 10:32

AllTooSir

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Some C++ compilers might flag it as unreachable as well. – James Kanze Aug 07 '13 at 11:44

score 1 · Answer 2 · answered Aug 07 '13 at 10:34

1

As per standard

Within nested statements, the break statement ends the do, for, switch, or while statement that immediately encloses it.

So in your case it will always break at first and never come to second break statement.

answered Aug 07 '13 at 10:34

PratAm

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score 0 · Answer 3 · answered Aug 07 '13 at 10:30

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The code at(1) will not be executed, break means jump out of the loop, so after the first break, the code will jump out of the inner loop.

answered Aug 07 '13 at 10:30

BitHigher

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score 0 · Answer 4 · answered Aug 07 '13 at 10:30

Within that particular loop, no, no code after break will execute as break will drop out of that loop and continue execution from after the end of the inner for loop. Although any code outside of the inner loop that appears after the break (within the outer loop), will be executed.

score 0 · Answer 5 · answered Aug 07 '13 at 10:30

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No, each time the first break statement is hit the inner loop will exit and the second break statement will never be reached.

answered Aug 07 '13 at 10:30

dunc123

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score 0 · Answer 6 · answered Aug 07 '13 at 10:31

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You can see here.

And I am sure the next break never exce . No matter what compiler you use.

answered Aug 07 '13 at 10:31

Lidong Guo

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score 0 · Answer 7 · answered Aug 07 '13 at 10:31

No. The code execution jumps out of the inner first loop the moment a break is encountered.
Even if the code is this way

break;
continue;
break;

the same thing happens. Continue is just a way to tell the compiler to iterate the loop further skipping any code in between.

Is the code after 'break' executed?

7 Answers7